Question

The velocity v of the flow of blood at a distance "r" from the central axis of an artery of radius "R" is: v = k(R^2^ - r^2^) where k is the constant of proportionality. Find the average rate of flow of blood along a radius of the artery. (use 0 and R as the limits of integration)

Answer 1

\[v=k(R^2-r^2)\]?

Answer 2

Yes, sorry

Answer 3

too damn many R's

Answer 4

are you supposed to integrate wrt
\[r\]?

Answer 5

I think I'm suppose to use the mean value thm....

Answer 6

well "average value" means integrate and divide by the length of the path

Answer 7

Im sorry, average value thm

Answer 8

and in this case they are telling you to integrate from 0 to R so length of path is just R

Answer 9

in other words you want
\[\frac{1}{R}\int_0^R k(R^2-r^2)dr\]

Answer 10

the R and r are not variables, right?

Answer 11

how would I apply (b - a) to this?

Answer 12

i think you are integrating wrt r

Answer 13

so in this case k, R are constants. b = R, a = 0 and b-a = R

Answer 14

i mean that is what it says in the problem you wrote. use 0 and R as the limits of inegration

Answer 15

when you integrate wrt r you get
\[\int _0^R kR^2-r^2 dr = \frac{2kR^3}{3}\]

Answer 16

and when you divide by R you get
\[\frac{2kR^2}{3}\]

Answer 17

here we are assuming that k is a constant, because it says
where k is the constant of proportionality.
and also R is a constant, the artery has radius R

Answer 18

what varies is the "flow of blood at distance r" at least that is the way i read the problem. the integral is relatively striaighforward

Answer 19

\[\int _0^R kR^2dr\] is just
\[R\times kR^2=kR^3\] because here R is constant

Answer 20

\[\int_0^R kr^2dr=\frac{kr^3}{3}|_0^R = \frac{kR^3}{3}\]

Answer 21

subtract and get
\[\int _0^R k(R^2-r^2)dr = \frac{2kR^3}{3}\]

Answer 22

giving
\[\frac{1}{R}\int_0^R k(R^2-r^2)dr = \frac{2kR^2}{3}\]

Answer 23

phew....... it seems to make sense. It's just going to a little while to sink in.... Thanks

Answer 24

hope i am right! yw

Answer 25

how would I apply (b - a) to this?

Answer 26

phew....... it seems to make sense. It's just going to a little while to sink in.... Thanks

Answer 27

I think you are